Computation of optimal stopping signals

3 Parameter-dependent optimal stopping In many applications, the reward function h of an optimal stopping problem depends on some parameter k, i.e. sup τ∈S E x ” e −R τ h k X τ — , x ∈ J , k ∈ R. A very prominent example is the American put option with strike k, i.e. h k x = k − x + . In the previous section, we have seen how to compute the corresponding value function for a fixed value of k by finding the smallest majorant of the reward function h k which is concave in some suitably generalized sense. An optimal stopping region Γ k is then given by the set where the value function and the reward function coincide cf. Proposition 6. If one has to determine optimal stopping rules corresponding to many different values of k, this approach might be quite tiresome, particularly if the structure of the stopping region is complex. In such a case, it would be desirable to have a universal stopping signal γ that characterizes the optimal stopping regions for any given parameter k, for instance as the level set Γ k = {x : γx ≥ k}. In the sequel, we describe a method to compute such a universal stopping signal for reward functions of the form h k x = ux − k, i.e. for the optimal stopping problem V k x ¬ sup τ∈S E x ” e −R τ uX τ − k — , x ∈ J , k ∈ R, 4 where u satisfies the following assumption: Assumption 4. The function u: J → R is continuously differentiable and lim sup x ↓a u + x ϕx ∞ and lim sup x ↑b u + x ψx ∞. Remark 7. In view of Proposition 6, Assumption 4 ensures that the value function V k of 4 is finite for any k.

3.1 Computation of optimal stopping signals

Let us discuss how to compute a universal stopping signal yielding an optimal stopping time for any k. To this end, let Γ k ¬ {x : V k x = ux − k} denote the optimal stopping region cf. Proposition 6. Assumption 5. For any k, the stopping time τ k ¬ inf {t : X t ∈ Γ k } is optimal, i.e. V k x = E x ” e −R τk uX τ k − k — . A sufficient condition for optimality of the stopping times τ k was given in Proposition 6. Moreover, granted there is an optimal stopping time, τ k is optimal as well see e.g. [El Karoui1981]. Remark 8. Recall that we use the convention hX τ = 0 on the set {τ = ∞} for any stopping time τ. The existence of an optimal stopping time τ k requires that the supremum in 4 is attained. 1976 We aim to derive a function γ on J such that the optimal stopping regions Γ k can be written as level sets Γ k = {x : γx ≥ k}. In other words, knowledge of the function γ suffices to derive the optimal stopping regions for any parameter k. Our main result, Theorem 13 below, shows that γx can be found as the infimum over an auxiliary function η x · of one real variable which is expressed in terms of the Laplace transforms ϕ and ψ of level passage times. The infimum can be computed explicitly under specific concavity conditions on the reward function u. In the sequel, we show how to obtain this result and give an explicit expression for η x in 12. Let T U ¬ inf {t ≥ 0 : X t 6∈ U} denote the first exit time from a measurable subset U of a, b and ˜ T x ¬ {T U : U ⊂ a, b open, x ∈ U} the class of all exit times from open neighborhoods of x. The following lemma shows that the stopping signal γ is the solution to a non-standard optimal stopping problem cf. equation 25 in [Bank and Föllmer2003], also [El Karoui and Föllmer2005]. Lemma 9. For γx ¬ inf T ∈ ˜ T x E x ” ux − e −R T uX T — E x 1 − e −R T , x ∈ J , 5 we have Γ k = {x : γx ≥ k}. Proof. For any k, we have x ∈ Γ k if and only if E x ” e −R T uX T − k — ≤ ux − k for any T ∈ ˜ T x. To see this, note that if x ∈ Γ k , then V k x = ux−k and the inequality above is true by definition of the value function. On the other hand, note that the sets Γ k are closed in J since V k is continuous for all k by Proposition 6. Thus, if x 6∈ Γ k , τ k = inf{t : X t ∈ Γ k } is an exit time from an open neighborhood of x, i.e. τ k ∈ ˜ T x. Since τ k is optimal by Assumption 5, we get ux − k V k x = E x ” e −R τk uX τ k − k — . Thus, for any k, we have x ∈ Γ k ⇐⇒ inf T ∈ ˜ T x E x ” ux − e −R T uX T — E x 1 − e −R T ≥ k. Let us now discuss how to compute the function γ of 5. The following lemma reduces the optimal stopping problem of the preceding lemma to finding the infimum of a function of two variables. Lemma 10. For any x ∈ J , γx = inf a ≤ yxz≤b f x y, z 6 1977 where f x y, z ¬                                ux −uz ψx ψz 1 − ψx ψz = ux ψx − uz ψz 1 ψx − 1 ψz , a = y x z b, ux ϕx − u y ϕ y F z −Fx F z −F y − uz ϕz F x −F y F z −F y 1 ϕx − 1 ϕ y F z −Fx F z −F y − 1 ϕz F x −F y F z −F y , a y x z b, ux −u y ϕx ϕ y 1 − ϕx ϕ y = ux ϕx − u y ϕ y 1 ϕx − 1 ϕ y , a y x z = b, ux y = a, z = b. Proof. First note that if T ∈ ˜ T x, T = T y ∧ T z P x -a.s. for some a ≤ y x z ≤ b. Thus, γx = inf a ≤ yxz≤b E x h ux − e −R T y ∧Tz uX T y ∧T z i E x ” 1 − e −R T y ∧Tz — Assumption 2 implies that T a = T b = ∞ a.s. The claim now follows from Lemma 1 and Lemma 3. Remark 11. According to Lemma 3, the functions ϕ and F may be replaced by ψ and G in the definition of f x , i.e. for a y x z b, f x y, z = ux ψx − u y ψ y · Gz −Gx Gz −G y − uz ψz · Gx −G y Gz −G y 1 ψx − 1 ψ y · Gz −Gx Gz −G y − 1 ψz · Gx −G y Gz −G y . For simplicity of notation, let α ϕ ¬ u ϕ , α ψ ¬ u ψ , β ϕ ¬ 1 ϕ , β ψ ¬ 1 ψ . For a y x z b, simple manipulations yield f x y, z = α ϕ x−α ϕ y F x −F y − α ϕ z−α ϕ x F z −Fx β ϕ x−β ϕ y F x −F y − β ϕ z−β ϕ x F z −Fx = α ψ x−α ψ y Gx −G y − α ψ z−α ψ x Gz −Gx β ψ x−β ψ y Gx −G y − β ψ z−β ψ x Gz −Gx . 7 We remark that f x y, z → f x a, z if y ↓ a iff u yϕ y → 0 as y ↓ a and f x y, z → f y, b if z ↑ b iff uzψz → 0 as z ↑ b. Let us discuss next how to compute the infimum of f x y, z over y and z for fixed x in order to find γx in 6. In view of the definition of f x , it seems natural to ask what happens if one lets y ↑ x or z ↓ x in 7. One notices that a difference quotient similar to the definition of the usual derivative appears. Let us therefore recall the following definition. 1978 Definition 12. Let f : J → R be a strictly monotone function. We say that U : J → R is f - differentiable at x ∈ J if d f Ux ¬ lim y →x U y − Ux f y − f x exists. For f s = s, the definition above is just the usual derivative. Also, if U and f are differentiable at x, then d f Ux = U ′ x f ′ x. In particular, under Assumption 4, α ϕ and α ψ are F - and G- differentiable. β ϕ and β ψ are always F - and G-differentiable. From 7, we find that lim y ↑x f x y, z = d F α ϕ x − α ϕ z−α ϕ x F z −Fx d F β ϕ x − β ϕ z−β ϕ x F z −Fx = d G α ψ x − α ψ z−α ψ x Gz −Gx d G β ψ x − β ψ z−β ψ x Gz −Gx , x z b, 8 lim z ↓x f x y, z = α ϕ x−α ϕ y F x −F y − d F α ϕ x β ϕ x−β ϕ y F x −F y − d F β ϕ x = α ψ x−α ψ y Gx −G y − d G α ψ x β ψ x−β ψ y Gx −G y − d G β ψ x , a y x. 9 Using l’Hôspital’s rule, one computes lim y ↑x f x y, b = ux − u ′ x ϕx ϕ ′ x = d F α ϕ x d F β ϕ x ¬ κx, 10 lim z ↓x f x a, z = ux − u ′ x ψx ψ ′ x = d G α ψ x d G β ψ x ¬ ρx. 11 We now define η x : [a, b] \ {x} → R by η x y ¬                ρx, y = a, d F α ϕ x− αϕ y−αϕ x F y −Fx d F β ϕ x− βϕ y−βϕ x F y −Fx , x y b, αϕ x−αϕ y F x −F y −d F α ϕ x βϕ x−βϕ y F x −F y −d F β ϕ x , a y x, κx, y = b. 12 η x · can also be written in terms of G, α ψ and β ψ instead of F, α ϕ and β ϕ in view of 8 and 9. In some proofs, it will be more convenient to use this alternative representation. Note that η x may fail to be continuous at the boundaries a or b. In fact, we have lim y ↓a η x y = ρx iff lim y ↓a u y ϕ y = 0 and lim z ↑b η x z = κx iff lim z ↑b uz ψz = 0. We now show that the computation of the optimal stopping signal γ at a point x ∈ J amounts to finding the infimum of the function η x of one real variable instead of the function f x of two variables. Theorem 13. Under Assumptions 1 - 5, for any x ∈ J = a, b, it holds that γx = inf y 6=x η x y, where η x is given by 12. 1979 In order to prove Theorem 13, we need the following lemma. Lemma 14. For any a ≤ y x z ≤ b, it holds that f x y, z ≥ min {η x y, η x z}. Proof. Suppose first that a y x z b. If f x y, z = 0, then either η x y ≤ 0 or η x z ≤ 0. Indeed, if both expressions were positive, then since f x y, z = 0, we would get considering only the numerators of f x in 7 and η x in 12 that 0 = α ϕ x − α ϕ y F x − F y − α ϕ z − α ϕ x F z − Fx = α ϕ x − α ϕ y F x − F y − d F α ϕ x + d F α ϕ x − α ϕ z − α ϕ x F z − Fx 0, a contradiction. If f x y, z 0, assume for the purpose of contradiction that the claim of the lemma is false, i.e. η x z = d F α ϕ x − α ϕ z−α ϕ x F z −Fx d F β ϕ x − β ϕ z−β ϕ x F z −Fx α ϕ x−α ϕ y F x −F y − α ϕ z−α ϕ x F z −Fx β ϕ x−β ϕ y F x −F y − β ϕ z−β ϕ x F z −Fx = f x y, z, η x y = α ϕ x−α ϕ y F x −F y − d F α ϕ x β ϕ x−β ϕ y F x −F y − d F β ϕ x α ϕ x−α ϕ y F x −F y − α ϕ z−α ϕ x F z −Fx β ϕ x−β ϕ y F x −F y − β ϕ z−β ϕ x F z −Fx = f x y, z. Strict F -concavity of β ϕ implies that all denominators are positive and the numerator of the right- hand side is also positive since f x y, z 0. Thus, we can rewrite the inequalities above as d F α ϕ x − α ϕ z−α ϕ x F z −Fx α ϕ x−α ϕ y F x −F y − α ϕ z−α ϕ x F z −Fx d F β ϕ x − β ϕ z−β ϕ x F z −Fx β ϕ x−β ϕ y F x −F y − β ϕ z−β ϕ x F z −Fx , α ϕ x−α ϕ y F x −F y − d F α ϕ x α ϕ x−α ϕ y F x −F y − α ϕ z−α ϕ x F z −Fx β ϕ x−β ϕ y F x −F y − d F β ϕ x β ϕ x−β ϕ y F x −F y − β ϕ z−β ϕ x F z −Fx . If we add both equations, we obtain 1 1, which is absurd. The case f x y, z 0 can be handled analogously. The proof in the case a y x, z = b is also along similar lines. Assume again that η x b = κx = d F α ϕ x d F β ϕ x α ϕ x − α ϕ y β ϕ x − β ϕ y = f x y, b, η x y = α ϕ x−α ϕ y F x −F y − d F α ϕ x β ϕ x−β ϕ y F x −F y − d F β ϕ x α ϕ x − α ϕ y β ϕ x − β ϕ y = f x y, b. 1980 If f x y, b = 0, then α ϕ x = α ϕ y, so we are led to the contradictory statement d F α ϕ x 0 and − d F α ϕ x 0. If f x y, b 6= 0, rearranging terms and adding the resulting equations as before, we also obtain a contradiction. If y = a and z b, assume again that the claim of the lemma is false, i.e. η x a = ρx = d G α ψ x d G β ψ x α ψ x − α ψ z β ψ x − β ψ z = f x a, z, η x z = d G α ψ x − α ψ z−α ψ x Gz −Gx d G β ψ x − β ψ z−β ψ x Gz −Gx α ψ x − α ψ z β ψ x − β ψ z = f x a, z. Here we have used the representation of η x involving G and ψ stated in 8. Since β ψ = 1ψ is G-concave and G is decreasing, this leads to a contradiction as before. Finally, if y = a, z = b, since ψ ′ 0 and ϕ ′ 0, clearly min {η x a, η x b} = min ux − u ′ x ψx ψ ′ x , ux − u ′ x ϕx ϕ ′ x ≤ ux. Proof. Theorem 13. By Lemma 10 and Lemma 14, it is immediate that for all x ∈ a, b γx = inf a ≤ yxz≤b f x y, z ≥ inf y 6=x η x y. To see the reverse inequality, note that for z x, η x z = lim u ↑x f x u, z ≥ inf a ≤ux f x u, z ≥ γx. Similarly, for y x, η x y = lim u ↓x f x y, u ≥ inf x u≤b f x y, u ≥ γx. Thus, inf y 6=x η x y ≥ γx.

3.2 Concave envelopes